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sergey [27]
3 years ago
5

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium

length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Physics
1 answer:
Kaylis [27]3 years ago
7 0

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

         w² = \frac{k}{m}

to find the constant (k) of the spring, we use Hooke's law with the initial data

         F = - kx

where the force is the weight of the body that is hanging

        F = W = m g

we substitute

        m g = - k x

        k = - \frac{m g}{x}

we calculate

        k = - \frac{9.8 m}{- 2.6 \ 10^{-2}}

        k = 3.769 10² m

we substitute in the first equation

       w² = \frac{ 3.769 \ 10^2 \ m }{m}

       w = 19.415 rad / s

angular velocity and frequency are related

       w = 2πf

        f = \frac{w}{2\pi }

        f = 19.415 / 2pi

        f = 3.09 Hz

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The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

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<u>Input our values:</u>

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<u>Divide both sides by 400 to isolate t:</u>

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<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>

<u></u>

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